One mole of an ideal gas with = (5/2)R and = (3/2)R expands from = 6 bar and = 800 K to = 1 bar by each of the following paths:

Question

(a) Constant volume
(b) Constant temperature
(c) Adiabatically
Assuming mechanical reversibility, calculate W, Q, ?U, and ?H for each process. Sketch each path on a single PV diagram.

Answer 1

For each of these problems start with equations



a. at constant volume,



So plugging the given values into each part gives:



Now not knowing what is, we must assume to stay at an ideal gas that since the pressure drops by a factor of 6, the temperature must drop by the same as well, leaving = 800/6 = 133.33 K. Using this value gives the following values:



b. Using the same starting point as in part (a) and assuming constant temperature, reduces the equations to the following:



Putting dQ and dW in terms of pressure and integrating gives:



Putting in the known values and solving gives



c. Again starting with the same equations as in part (a) and assume dQ = 0 (adiabatic) gives:



However, we also have adiabatic expansion which changes integrating and solving for gives:



Plugging in given values and solving gives: